Asymptotic gain model

The asymptotic gain model [Middlebrook, RD: "Design-oriented analysis of feedback amplifiers"; Proc. of National Electronics Conference, Vol. XX, Oct. 1964, pp. 1-4] G_{infin} T} .

while in classical feedback theory, in terms of the open loop gain "A", the gain with feedback (closed loop gain) is:

::

Comparison of the two expressions indicates the feedback factor β_FB is:

::

while the open-loop gain is:

::$A\; =\; G\_\{infin\}\; T\; .$

If the accuracy is adequate (usually it is), these formulas suggest an alternative evaluation of "T": evaluate the open-loop gain and "G_∞" and use these expressions to find "T". Often these two evaluations are easier than evaluation of "T" directly.

Examples

The steps in deriving the gain using the asymptotic gain formula are outlined below for two negative feedback amplifiers. The single transistor example shows how the method works in principle for a transconductance amplifier, while the second two-transistor example shows the approach to more complex cases using a current amplifier.

ingle-stage transistor amplifier

Consider the simple FET feedback amplifier in Figure 3. The aim is to find the low-frequency, open-circuit, transresistance gain of this circuit "G" = "v_out" / "i _in" using the asymptotic gain model.

The small-signal equivalent circuit is shown in Figure 4, where the transistor is replaced by its hybrid-pi model.

Return ratio

It is most straightforward to begin by finding the return ratio "T", because "G₀" and "G_∞" are defined as limiting forms of the gain as "T" tends to either zero or infinity. To take these limits, it is necessary to know what parameters "T" depends upon. There is only one dependent source in this circuit, so as a starting point the return ratio related to this source is determined as outlined in the article on return ratio.

The return ratio is found using Figure 5. In Figure 5, the input current source is set to zero, By cutting the dependent source out of the output side of the circuit, and short-circuiting its terminals, the output side of the circuit is isolated from the input and the feedback loop is broken. A test current "i_t" replaces the dependent source. Then the return current generated in the dependent source by the test current is found. The return ratio is then "T" = −"i_r / i_t". Using this method, and noticing that "R_D" is in parallel with "r_O", "T" is determined as::$T\; =\; g\_m\; left(\; R\_D\; ||r\_O\; ight)\; approx\; g\_m\; R\_D\; ,$where the approximation is accurate in the common case where "r_O" >> "R_D". With this relationship it is clear that the limits "T" → 0, or ∞ are realized if we let transconductance "g_m" → 0, or ∞. [Although changing "R_D // r_O" also could force the return ratio limits, these resistor values affect other aspects of the circuit as well. It is the "control parameter" of the dependent source that must be varied because it affects "only" the dependent source.]

Asymptotic gain

Finding the asymptotic gain "G_∞" provides insight, and usually can be done by inspection. To find "G_∞" we let "g_m" → ∞ and find the resulting gain. The drain current, "i_D" = "g_m" "v_GS", must be finite. Hence, as "g_m" approaches infinity, "v_GS" also must approach zero. As the source is grounded, "v_GS" = 0 implies "v_G" = 0 as well. [Because the input voltage "v_GS" approaches zero as the return ratio gets larger, the amplifier input impedance also tends to zero, which means in turn (because of current division) that the amplifier works best if the input signal is a current. If a Norton source is used, rather than an ideal current source, the formal equations derived for "T" will be the same as for a Thévenin voltage source. Note that in the case of input current, "G_∞" is a transresistance gain.] With "v_G" = 0 and the fact that all the input current flows through "R_f" (as the FET has an infinite input impedance), the output voltage is simply −"i_in R_f". Hence

:$G\_\{infty\}\; =\; frac\{v\_\{out\{i\_\{in\; =\; -R\_f\; .$

Alternatively "G_∞" is the gain found by replacing the transistor by an ideal amplifier with infinite gain - a nullor.] That is, the amplifier uses current feedback. It frequently is ambiguous just what type of feedback is involved in an amplifier, and the asymptotic gain approach has the advantage/disadvantage that it works whether or not you understand the circuit.

Figure 6 indicates the output node, but does not indicate the choice of output variable. In what follows, the output variable is selected as the short-circuit current of the amplifier, that is, the collector current of the output transistor. Other choices for output are discussed later.

To implement the asymptotic gain model, the dependent source associated with either transistor can be used. Here the first transistor is chosen.

Return ratio

The circuit to determine the return ratio is shown in the top panel of Figure 7. Labels show the currents in the various branches as found using a combination of Ohm's law and Kirchhoff's laws. Resistor "R₁ = R_B // r_π1" and "R₃ = R_C2 // R_L". KVL from the ground of "R₁" to the ground of "R₂" provides:

:

KVL provides the collector voltage at the top of "R_C" as

:$v\_C\; =\; v\_\{\; pi\}\; left(1+\; frac\; \{R\_f\}\; \{R\_1\}\; ight\; )\; -i\_B\; r\_\{\; pi\; 2\}\; .$

Finally, KCL at this collector provides

:$i\_T\; =\; i\_B\; -\; frac\; \{v\_C\}\; \{R\_\{C\; .$

Substituting the first equation into the second and the second into the third, the return ratio is found as

:$T\; =\; -\; frac\; \{i\_R\}\; \{i\_T\}\; =\; -g\_m\; frac\; \{v\_\{\; pi\}\; \}\{i\_T\}$:::

=Gain "G₀" with T = 0 =

The circuit to determine "G₀" is shown in the center panel of Figure 7. In Figure 7, the output variable is the output current β"i_B" (the short-circuit load current), which leads to the short-circuit current gain of the amplifier, namely β"i_B" / "i"_S:

::

Using Ohm's law, the voltage at the top of "R₁" is found as

::$(\; i\_S\; -\; i\_R\; )\; R\_1\; =\; i\_R\; R\_f\; +v\_E\; ,$

or, rearranging terms,

::$i\_S\; =\; i\_R\; left(\; 1\; +\; frac\; \{R\_f\}\{R\_1\}\; ight)\; +frac\; \{v\_E\}\; \{R\_1\}\; .$

Using KCL at the top of "R₂":

::

Emitter voltage "v_E" already is known in terms of "i_B" from the diagram of Figure 7. Substituting the second equation in the first, "i_B" is determined in terms of "i_S" alone, and "G₀" becomes:

::

Gain "G₀" represents feedforward through the feedback network, and commonly is negligible.

Gain "G_∞" with "T" → ∞

The circuit to determine "G_∞" is shown in the bottom panel of Figure 7. The introduction of the ideal op amp (a nullor) in this circuit is explained as follows. When "T "→ ∞, the gain of the amplifier goes to infinity as well, and in such a case the differential voltage driving the amplifier (the voltage across the input transistor "r_π1") is driven to zero and (according to Ohm's law when there is no voltage) it draws no input current. On the other hand the output current and output voltage are whatever the circuit demands. This behavior is like a nullor, so a nullor can be introduced to represent the infinite gain transistor.

The current gain is read directly off the schematic:

::

Comparison with classical feedback theory

Using the classical model, the feed-forward is neglected and the feedback factor β_FB is (assuming transistor β >> 1):

::

and the open-loop gain "A" is:

::

Overall gain

The above expressions can be substituted into the asymptotic gain model equation to find the overall gain G. The resulting gain is the "current" gain of the amplifier with a short-circuit load.

Gain using alternative output variables

In the amplifier of Figure 6, "R_L" and "R_C2" are in parallel.To obtain the transresistance gain, say "A"_ρ, that is, the gain using voltage as output variable, the short-circuit current gain "G" is multiplied by "R_C2 // R_L" in accordance with Ohm's law:

::$A\_\{\; ho\}\; =\; G\; left(\; R\_\{C2\}\; //\; R\_\{L\}\; ight)\; .$

The "open-circuit" voltage gain is found from "A"_ρ by setting "R"_L → ∞. To obtain the current gain when load current "i_L" in load resistor "R"_L is the output variable, say "A"_i, the formula for current division is used: "i_L = i_out × R_C2 / ( R_C2 + R_L )" and the short-circuit current gain "G" is multiplied by this loading factor:

::$A\_i\; =\; G\; left(\; frac\; \{R\_\{C2\; \{R\_\{C2\}+\; R\_\{L\; ight)\; .$

Of course, the short-circuit current gain is recovered by setting "R"_L = 0 Ω.

References and notes

ee also

*Return ratio
*Signal-flow graph
*Feedback amplifiers

Links

* [http://users.ece.gatech.edu/~pallen/Academic/ECE_6412/Spring_2004/L290-ReturnRatio-2UP.pdf Lecture notes on the asymptotic gain model]